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To solve the given vector differential equation [tex]\( \mathbf{r}'(t) = 8t^{-1} \mathbf{i} + 10t \mathbf{j} + 21t^2 \mathbf{k} \)[/tex] with the initial condition [tex]\( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex], follow these steps:
1. Write the differential equation in component form:
[tex]\[ \mathbf{r}'(t) = \left(8t^{-1}\right)\mathbf{i} + \left(10t\right)\mathbf{j} + \left(21t^2\right)\mathbf{k} \][/tex]
2. Integrate each component of the vector:
- For the [tex]\(\mathbf{i}\)[/tex]-component:
[tex]\[ \int 8t^{-1} \, dt = 8 \int t^{-1} \, dt = 8 \ln |t| + C_1 \][/tex]
Since [tex]\(t > 0\)[/tex], we can write the answer as:
[tex]\[ 8 \ln t + C_1 \][/tex]
- For the [tex]\(\mathbf{j}\)[/tex]-component:
[tex]\[ \int 10t \, dt = 10 \int t \, dt = 10 \left(\frac{t^2}{2}\right) + C_2 = 5t^2 + C_2 \][/tex]
- For the [tex]\(\mathbf{k}\)[/tex]-component:
[tex]\[ \int 21t^2 \, dt = 21 \int t^2 \, dt = 21 \left(\frac{t^3}{3}\right) + C_3 = 7t^3 + C_3 \][/tex]
3. Combine the integrated components to form the general solution of the vector [tex]\( \mathbf{r}(t) \)[/tex]:
[tex]\[ \mathbf{r}(t) = (8 \ln t + C_1)\mathbf{i} + (5t^2 + C_2)\mathbf{j} + (7t^3 + C_3)\mathbf{k} \][/tex]
4. Apply the initial condition [tex]\( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex]:
[tex]\[ \mathbf{r}(1) = (8 \ln 1 + C_1)\mathbf{i} + (5 \cdot 1^2 + C_2)\mathbf{j} + (7 \cdot 1^3 + C_3)\mathbf{k} \][/tex]
Simplify using [tex]\( \ln 1 = 0 \)[/tex]:
[tex]\[ \mathbf{r}(1) = C_1 \mathbf{i} + (5 + C_2) \mathbf{j} + (7 + C_3) \mathbf{k} \][/tex]
Set this equal to [tex]\( \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex]:
[tex]\[ C_1 \mathbf{i} + (5 + C_2) \mathbf{j} + (7 + C_3) \mathbf{k} = \mathbf{i} + \mathbf{j} + \mathbf{k} \][/tex]
This gives us the equations for the constants [tex]\( C_1, C_2, \)[/tex] and [tex]\( C_3 \)[/tex]:
[tex]\[ C_1 = 1 \][/tex]
[tex]\[ 5 + C_2 = 1 \implies C_2 = -4 \][/tex]
[tex]\[ 7 + C_3 = 1 \implies C_3 = -6 \][/tex]
5. Substitute the constants [tex]\( C_1, C_2, \)[/tex] and [tex]\( C_3 \)[/tex] back into the general solution:
[tex]\[ \mathbf{r}(t) = (8 \ln t + 1) \mathbf{i} + (5t^2 - 4) \mathbf{j} + (7t^3 - 6) \mathbf{k} \][/tex]
Thus, the solution to the vector differential equation [tex]\( \mathbf{r}'(t) = 8t^{-1} \mathbf{i} + 10t \mathbf{j} + 21t^2 \mathbf{k} \)[/tex] with the initial condition [tex]\( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex] is:
[tex]\[ \mathbf{r}(t) = (8 \ln t + 1) \mathbf{i} + (5t^2 - 4) \mathbf{j} + (7t^3 - 6) \mathbf{k} \][/tex]
1. Write the differential equation in component form:
[tex]\[ \mathbf{r}'(t) = \left(8t^{-1}\right)\mathbf{i} + \left(10t\right)\mathbf{j} + \left(21t^2\right)\mathbf{k} \][/tex]
2. Integrate each component of the vector:
- For the [tex]\(\mathbf{i}\)[/tex]-component:
[tex]\[ \int 8t^{-1} \, dt = 8 \int t^{-1} \, dt = 8 \ln |t| + C_1 \][/tex]
Since [tex]\(t > 0\)[/tex], we can write the answer as:
[tex]\[ 8 \ln t + C_1 \][/tex]
- For the [tex]\(\mathbf{j}\)[/tex]-component:
[tex]\[ \int 10t \, dt = 10 \int t \, dt = 10 \left(\frac{t^2}{2}\right) + C_2 = 5t^2 + C_2 \][/tex]
- For the [tex]\(\mathbf{k}\)[/tex]-component:
[tex]\[ \int 21t^2 \, dt = 21 \int t^2 \, dt = 21 \left(\frac{t^3}{3}\right) + C_3 = 7t^3 + C_3 \][/tex]
3. Combine the integrated components to form the general solution of the vector [tex]\( \mathbf{r}(t) \)[/tex]:
[tex]\[ \mathbf{r}(t) = (8 \ln t + C_1)\mathbf{i} + (5t^2 + C_2)\mathbf{j} + (7t^3 + C_3)\mathbf{k} \][/tex]
4. Apply the initial condition [tex]\( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex]:
[tex]\[ \mathbf{r}(1) = (8 \ln 1 + C_1)\mathbf{i} + (5 \cdot 1^2 + C_2)\mathbf{j} + (7 \cdot 1^3 + C_3)\mathbf{k} \][/tex]
Simplify using [tex]\( \ln 1 = 0 \)[/tex]:
[tex]\[ \mathbf{r}(1) = C_1 \mathbf{i} + (5 + C_2) \mathbf{j} + (7 + C_3) \mathbf{k} \][/tex]
Set this equal to [tex]\( \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex]:
[tex]\[ C_1 \mathbf{i} + (5 + C_2) \mathbf{j} + (7 + C_3) \mathbf{k} = \mathbf{i} + \mathbf{j} + \mathbf{k} \][/tex]
This gives us the equations for the constants [tex]\( C_1, C_2, \)[/tex] and [tex]\( C_3 \)[/tex]:
[tex]\[ C_1 = 1 \][/tex]
[tex]\[ 5 + C_2 = 1 \implies C_2 = -4 \][/tex]
[tex]\[ 7 + C_3 = 1 \implies C_3 = -6 \][/tex]
5. Substitute the constants [tex]\( C_1, C_2, \)[/tex] and [tex]\( C_3 \)[/tex] back into the general solution:
[tex]\[ \mathbf{r}(t) = (8 \ln t + 1) \mathbf{i} + (5t^2 - 4) \mathbf{j} + (7t^3 - 6) \mathbf{k} \][/tex]
Thus, the solution to the vector differential equation [tex]\( \mathbf{r}'(t) = 8t^{-1} \mathbf{i} + 10t \mathbf{j} + 21t^2 \mathbf{k} \)[/tex] with the initial condition [tex]\( \mathbf{r}(1) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex] is:
[tex]\[ \mathbf{r}(t) = (8 \ln t + 1) \mathbf{i} + (5t^2 - 4) \mathbf{j} + (7t^3 - 6) \mathbf{k} \][/tex]
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